# Find the maximum volume of an open box made from 3ft by 8ft rectangular piece of sheet metal by cutting out squares of equal sizes of four corners

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Find the maximum volume of an open box made from a 3' by 8' rectangular sheet with equal squares cut fromt he 4 corners:

The volume is `V=(3-2x)(8-2x)x`

** Each length has a segment of length x cut from both corners; the height of the box will be x **

To find the maximum we take the derivative with respect to x.

`V=4x^3-22x^2+24x` after expanding so:

`(dV)/(dx)=12x^2-44x+24=4(3x^2-11x+6)=4(3x-2)(x-3)`

A maximum of a continuous function can only occur at an critical point (where `f'(x)=0` or `f'(x)` fails to exist. Since the derivative is a polynomial it is everywhere continuous).

Setting `(dV)/(dx)=0` we get `4(3x-2)(x-3)=0==>x=2/3` or `x=3`

`x=3` does not make sense in the context of the problem. (After cutting squares of length 3 from the side of length 3, we have a negative side length) so x=`2/3` .

If x=`2/3` the volume of the box is `V=(3-4/3)(8-4/3)(2/3)=200/27`

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**The maximum volume is `200/27~~7.41` cu ft**

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